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Unformatted text preview: is correct. P (correct order) = 1 4! = 1 24 . 6. This problem is similar to problem 4. First assume all the 7 heads and 3 tails are distinct. Then there are 10! ways. But the 7 heads are identical and the 3 tails are identical. Thus, the total number of ways is 10! / (7!3!) = 120. 7. Combinations of 1 out of 5 = ‡ 5 1 · = 5; combinations of 2 out of 5 = ‡ 5 2 · = 5! / 3!2! = 10; sum of all combinations = ∑ 5 j =0 ‡ 5 j · = 2 5 = 32. 8. Combinations of 5 out of 20 = ‡ 20 5 · . The number of ways of selecting 2 defectives out of 3 defectives and 3 nondefectives out of 17 nondefectives = ‡ 3 3 ·‡ 17 3 · . Thus, P2 defectives in 5 samples = ‡ 3 2 ·‡ 17 3 · / ‡ 20 5 · ....
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This note was uploaded on 06/09/2010 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.
 Spring '08
 LORENZELLI

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